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Groups > sci.physics.relativity > #380413 > unrolled thread
| Started by | Julio Di Egidio <julio@diegidio.name> |
|---|---|
| First post | 2016-03-29 04:40 -0700 |
| Last post | 2016-03-31 07:20 -0700 |
| Articles | 20 on this page of 114 — 13 participants |
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Re: Inescapable (symmetric) twins paradox Julio Di Egidio <julio@diegidio.name> - 2016-03-29 04:40 -0700
Re: Inescapable (symmetric) twins paradox "Paul B. Andersen" <relativity@paulba.no> - 2016-03-29 14:54 +0200
Re: Inescapable (symmetric) twins paradox Julio Di Egidio <julio@diegidio.name> - 2016-04-03 15:13 -0700
Re: Inescapable (symmetric) twins paradox moroney@world.std.spaamtrap.com (Michael Moroney) - 2016-04-04 01:03 +0000
Re: Inescapable (symmetric) twins paradox David Waite <waitedavid1618@yahoo.com> - 2016-04-03 20:57 -0700
Re: Inescapable (symmetric) twins paradox Koobee Wublee <koobee.wublee@gmail.com> - 2016-04-03 22:14 -0700
Re: Inescapable (symmetric) twins paradox David Waite <waitedavid1618@yahoo.com> - 2016-04-04 10:53 -0700
Re: Inescapable (symmetric) twins paradox David Waite <waitedavid1618@yahoo.com> - 2016-04-04 11:24 -0700
Re: Inescapable (symmetric) twins paradox Koobee Wublee <koobee.wublee@gmail.com> - 2016-04-04 23:44 -0700
Re: Inescapable (symmetric) twins paradox David Waite <waitedavid1618@yahoo.com> - 2016-04-05 02:05 -0700
Re: Inescapable (symmetric) twins paradox mlwozniak@wp.pl - 2016-04-05 02:51 -0700
Re: Inescapable (symmetric) twins paradox David Waite <waitedavid1618@yahoo.com> - 2016-04-05 02:58 -0700
Re: Inescapable (symmetric) twins paradox mlwozniak@wp.pl - 2016-04-05 04:08 -0700
Re: Inescapable (symmetric) twins paradox David Waite <waitedavid1618@yahoo.com> - 2016-04-05 04:25 -0700
Re: Inescapable (symmetric) twins paradox mlwozniak@wp.pl - 2016-04-05 04:41 -0700
Re: Inescapable (symmetric) twins paradox David Waite <waitedavid1618@yahoo.com> - 2016-04-05 05:14 -0700
Re: Inescapable (symmetric) twins paradox mlwozniak@wp.pl - 2016-04-05 05:46 -0700
Re: Inescapable (symmetric) twins paradox David Waite <waitedavid1618@yahoo.com> - 2016-04-05 06:46 -0700
Re: Inescapable (symmetric) twins paradox Maciej Woźniak <mlwozniak@wp.pl> - 2016-04-05 20:54 +0200
Re: Inescapable (symmetric) twins paradox David Waite <waitedavid1618@yahoo.com> - 2016-04-05 12:45 -0700
Re: Inescapable (symmetric) twins paradox mlwozniak@wp.pl - 2016-04-05 23:17 -0700
Re: Inescapable (symmetric) twins paradox alsor@interia.pl - 2016-04-06 06:41 -0700
Re: Inescapable (symmetric) twins paradox David Waite <waitedavid1618@yahoo.com> - 2016-04-06 07:56 -0700
Re: Inescapable (symmetric) twins paradox alsor@interia.pl - 2016-04-06 09:00 -0700
Re: Inescapable (symmetric) twins paradox David Waite <waitedavid1618@yahoo.com> - 2016-04-06 09:54 -0700
Re: Inescapable (symmetric) twins paradox alsor@interia.pl - 2016-04-06 11:51 -0700
Re: Inescapable (symmetric) twins paradox JanPB <filmart@gmail.com> - 2016-04-06 11:29 -0700
Re: Inescapable (symmetric) twins paradox alsor@interia.pl - 2016-04-06 12:01 -0700
Re: Inescapable (symmetric) twins paradox JanPB <filmart@gmail.com> - 2016-04-06 14:33 -0700
Re: Inescapable (symmetric) twins paradox alsor@interia.pl - 2016-04-09 10:17 -0700
Re: Inescapable (symmetric) twins paradox JanPB <filmart@gmail.com> - 2016-04-09 15:45 -0700
Re: Inescapable (symmetric) twins paradox alsor@interia.pl - 2016-04-10 12:08 -0700
Re: Inescapable (symmetric) twins paradox mlwozniak@wp.pl - 2016-04-03 23:48 -0700
Re: Inescapable (symmetric) twins paradox "Paul B. Andersen" <relativity@paulba.no> - 2016-04-04 20:22 +0200
Re: Inescapable (symmetric) twins paradox "Dono," <sa_ge@comcast.net> - 2016-03-29 06:45 -0700
Re: Inescapable (symmetric) twins paradox David Waite <waitedavid1618@yahoo.com> - 2016-03-29 08:18 -0700
Re: Inescapable (symmetric) twins paradox alsor@interia.pl - 2016-03-29 11:17 -0700
Re: Inescapable (symmetric) twins paradox David Waite <waitedavid1618@yahoo.com> - 2016-03-29 11:54 -0700
Re: Inescapable (symmetric) twins paradox Maciej Woźniak <mlwozniak@wp.pl> - 2016-03-29 21:50 +0200
Re: Inescapable (symmetric) twins paradox Maciej Woźniak <mlwozniak@wp.pl> - 2016-03-29 22:09 +0200
Re: Inescapable (symmetric) twins paradox David Waite <waitedavid1618@yahoo.com> - 2016-03-29 13:11 -0700
Re: Inescapable (symmetric) twins paradox Maciej Woźniak <mlwozniak@wp.pl> - 2016-03-29 22:43 +0200
Re: Inescapable (symmetric) twins paradox David Waite <waitedavid1618@yahoo.com> - 2016-03-29 14:24 -0700
Re: Inescapable (symmetric) twins paradox JanPB <filmart@gmail.com> - 2016-03-29 15:58 -0700
Re: Inescapable (symmetric) twins paradox mlwozniak@wp.pl - 2016-03-29 23:15 -0700
Re: Inescapable (symmetric) twins paradox JanPB <filmart@gmail.com> - 2016-03-30 00:07 -0700
Re: Inescapable (symmetric) twins paradox mlwozniak@wp.pl - 2016-03-30 00:51 -0700
Re: Inescapable (symmetric) twins paradox David Waite <waitedavid1618@yahoo.com> - 2016-03-30 01:45 -0700
Re: Inescapable (symmetric) twins paradox mlwozniak@wp.pl - 2016-03-30 02:06 -0700
Re: Inescapable (symmetric) twins paradox David Waite <waitedavid1618@yahoo.com> - 2016-03-30 02:35 -0700
Re: Inescapable (symmetric) twins paradox mlwozniak@wp.pl - 2016-03-30 03:00 -0700
Re: Inescapable (symmetric) twins paradox David Waite <waitedavid1618@yahoo.com> - 2016-03-30 03:11 -0700
Re: Inescapable (symmetric) twins paradox mlwozniak@wp.pl - 2016-03-30 03:21 -0700
Re: Inescapable (symmetric) twins paradox David Waite <waitedavid1618@yahoo.com> - 2016-03-30 03:43 -0700
Re: Inescapable (symmetric) twins paradox mlwozniak@wp.pl - 2016-03-30 03:53 -0700
Re: Inescapable (symmetric) twins paradox David Waite <waitedavid1618@yahoo.com> - 2016-03-30 04:34 -0700
Re: Inescapable (symmetric) twins paradox Julio Di Egidio <julio@diegidio.name> - 2016-03-30 04:43 -0700
Re: Inescapable (symmetric) twins paradox mlwozniak@wp.pl - 2016-03-30 05:31 -0700
Re: Inescapable (symmetric) twins paradox David Waite <waitedavid1618@yahoo.com> - 2016-03-30 06:08 -0700
Re: Inescapable (symmetric) twins paradox JanPB <filmart@gmail.com> - 2016-03-30 12:21 -0700
Re: Inescapable (symmetric) twins paradox Odd Bodkin <bodkinodd@gmail.com> - 2016-03-30 14:48 -0500
Re: Inescapable (symmetric) twins paradox mlwozniak@wp.pl - 2016-03-30 23:39 -0700
Re: Inescapable (symmetric) twins paradox Maciej Woźniak <mlwozniak@wp.pl> - 2016-03-30 22:34 +0200
Re: Inescapable (symmetric) twins paradox JanPB <filmart@gmail.com> - 2016-03-30 14:36 -0700
Re: Inescapable (symmetric) twins paradox mlwozniak@wp.pl - 2016-03-30 23:17 -0700
Re: Inescapable (symmetric) twins paradox JanPB <filmart@gmail.com> - 2016-03-31 00:04 -0700
Re: Inescapable (symmetric) twins paradox mlwozniak@wp.pl - 2016-03-31 00:15 -0700
Re: Inescapable (symmetric) twins paradox JanPB <filmart@gmail.com> - 2016-03-31 00:30 -0700
Re: Inescapable (symmetric) twins paradox mlwozniak@wp.pl - 2016-03-30 05:29 -0700
Re: Inescapable (symmetric) twins paradox JanPB <filmart@gmail.com> - 2016-03-30 12:13 -0700
Re: Inescapable (symmetric) twins paradox Maciej Woźniak <mlwozniak@wp.pl> - 2016-03-30 22:30 +0200
Re: Inescapable (symmetric) twins paradox JanPB <filmart@gmail.com> - 2016-03-30 13:35 -0700
Re: Inescapable (symmetric) twins paradox Maciej Woźniak <mlwozniak@wp.pl> - 2016-03-30 22:53 +0200
Re: Inescapable (symmetric) twins paradox JanPB <filmart@gmail.com> - 2016-03-30 14:39 -0700
Re: Inescapable (symmetric) twins paradox mlwozniak@wp.pl - 2016-03-30 23:23 -0700
Re: Inescapable (symmetric) twins paradox JanPB <filmart@gmail.com> - 2016-03-31 00:06 -0700
Re: Inescapable (symmetric) twins paradox mlwozniak@wp.pl - 2016-03-31 00:23 -0700
Re: Inescapable (symmetric) twins paradox JanPB <filmart@gmail.com> - 2016-03-31 00:42 -0700
Re: Inescapable (symmetric) twins paradox mlwozniak@wp.pl - 2016-03-31 00:56 -0700
Re: Inescapable (symmetric) twins paradox JanPB <filmart@gmail.com> - 2016-03-31 11:11 -0700
Re: Inescapable (symmetric) twins paradox Maciej Woźniak <mlwozniak@wp.pl> - 2016-03-31 22:10 +0200
Re: Inescapable (symmetric) twins paradox JanPB <filmart@gmail.com> - 2016-03-31 13:50 -0700
Re: Inescapable (symmetric) twins paradox Maciej Woźniak <mlwozniak@wp.pl> - 2016-03-31 23:03 +0200
Re: Inescapable (symmetric) twins paradox JanPB <filmart@gmail.com> - 2016-03-31 15:01 -0700
Re: Inescapable (symmetric) twins paradox mlwozniak@wp.pl - 2016-03-31 23:09 -0700
Re: Inescapable (symmetric) twins paradox JanPB <filmart@gmail.com> - 2016-04-01 00:11 -0700
Re: Inescapable (symmetric) twins paradox mlwozniak@wp.pl - 2016-04-01 00:34 -0700
Re: Inescapable (symmetric) twins paradox moroney@world.std.spaamtrap.com (Michael Moroney) - 2016-04-01 14:53 +0000
Re: Inescapable (symmetric) twins paradox Maciej Woźniak <mlwozniak@wp.pl> - 2016-04-01 17:55 +0200
Re: Inescapable (symmetric) twins paradox moroney@world.std.spaamtrap.com (Michael Moroney) - 2016-04-01 18:44 +0000
Re: Inescapable (symmetric) twins paradox Maciej Woźniak <mlwozniak@wp.pl> - 2016-04-01 21:02 +0200
Re: Inescapable (symmetric) twins paradox moroney@world.std.spaamtrap.com (Michael Moroney) - 2016-04-04 01:10 +0000
Re: Inescapable (symmetric) twins paradox JanPB <filmart@gmail.com> - 2016-04-01 10:30 -0700
Re: Inescapable (symmetric) twins paradox alsor@interia.pl - 2016-04-01 10:38 -0700
Re: Inescapable (symmetric) twins paradox JanPB <filmart@gmail.com> - 2016-04-01 13:36 -0700
Re: Inescapable (symmetric) twins paradox alsor@interia.pl - 2016-04-04 11:41 -0700
Re: Inescapable (symmetric) twins paradox JanPB <filmart@gmail.com> - 2016-04-04 15:02 -0700
Re: Inescapable (symmetric) twins paradox mlwozniak@wp.pl - 2016-04-04 23:01 -0700
Re: Inescapable (symmetric) twins paradox JanPB <filmart@gmail.com> - 2016-04-05 00:27 -0700
Re: Inescapable (symmetric) twins paradox mlwozniak@wp.pl - 2016-04-05 00:52 -0700
Re: Inescapable (symmetric) twins paradox Gary Harnagel <hitlong@yahoo.com> - 2016-04-01 15:37 -0700
Re: Inescapable (symmetric) twins paradox Maciej Woźniak <mlwozniak@wp.pl> - 2016-04-01 21:10 +0200
Re: Inescapable (symmetric) twins paradox Bimpy <pimp@pimp.com> - 2016-04-01 19:15 +0000
Re: Inescapable (symmetric) twins paradox JanPB <filmart@gmail.com> - 2016-04-01 13:38 -0700
Re: Inescapable (symmetric) twins paradox Maciej Woźniak <mlwozniak@wp.pl> - 2016-04-02 09:05 +0200
Re: Inescapable (symmetric) twins paradox moroney@world.std.spaamtrap.com (Michael Moroney) - 2016-03-31 14:42 +0000
Re: Inescapable (symmetric) twins paradox David Waite <waitedavid1618@yahoo.com> - 2016-03-30 02:45 -0700
Re: Inescapable (symmetric) twins paradox JanPB <filmart@gmail.com> - 2016-03-30 12:10 -0700
Re: Inescapable (symmetric) twins paradox JanPB <filmart@gmail.com> - 2016-03-30 12:07 -0700
Re: Inescapable (symmetric) twins paradox Maciej Woźniak <mlwozniak@wp.pl> - 2016-03-30 22:26 +0200
Re: Inescapable (symmetric) twins paradox mlwozniak@wp.pl - 2016-03-29 23:25 -0700
Re: Inescapable (symmetric) twins paradox alsor@interia.pl - 2016-03-30 10:30 -0700
Re: Inescapable (symmetric) twins paradox David Waite <waitedavid1618@yahoo.com> - 2016-03-30 10:48 -0700
Re: Inescapable (symmetric) twins paradox alsor@interia.pl - 2016-03-31 07:20 -0700
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| From | Julio Di Egidio <julio@diegidio.name> |
|---|---|
| Date | 2016-03-29 04:40 -0700 |
| Subject | Re: Inescapable (symmetric) twins paradox |
| Message-ID | <1814ad0d-bcd0-4e58-b114-e80930174260@googlegroups.com> |
On Wednesday, July 15, 2015 at 3:55:53 PM UTC+1, Julio Di Egidio wrote: > Dear all, > > "In the context of special relativity, we present a twins experiment that is > symmetric between the twins, so that a paradox appears inescapable, in the > form of a violation of the principle of causality. [...] Bottom line: if > special relativity is correct, it must be incomplete." > > <http://seprogrammo.blogspot.co.uk/2015/07/symmetric-twins-paradox.html> Here is another little "experiment" (requires a recent browser): <https://jsfiddle.net/juliopdiegidio/aqxfLvs2/embedded/result/> It is a symmetric setup, only the radial motion is considered. The first diagram and figures shows the point of view of a C that stays in the mid-point between A and B. The second diagram and figures shows the point of view of A. The two points of view (of C and of A) turn out to be incompatible: there is a rendez-vous between all three observes from the point of view of C, but no such rendez-vous from the point of view of A (or, symmetrically, for B). -- But I might also have made some basic mistake. Anybody here knows how to fix it ?? Julio
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| From | "Paul B. Andersen" <relativity@paulba.no> |
|---|---|
| Date | 2016-03-29 14:54 +0200 |
| Message-ID | <nddtue$373$1@news.albasani.net> |
| In reply to | #380413 |
On 29.03.2016 13:40, Julio Di Egidio wrote: > On Wednesday, July 15, 2015 at 3:55:53 PM UTC+1, Julio Di Egidio wrote: >> Dear all, >> >> "In the context of special relativity, we present a twins experiment that is >> symmetric between the twins, so that a paradox appears inescapable, in the >> form of a violation of the principle of causality. [...] Bottom line: if >> special relativity is correct, it must be incomplete." >> >> <http://seprogrammo.blogspot.co.uk/2015/07/symmetric-twins-paradox.html> Sorry, you are simply wrong. If you apply the special relativity (SR) on your scenario, you will find that it predicts that the clocks will show the same when the twins are reunited. Your guesses of what SR predicts are wrong! Do the math! > > Here is another little "experiment" (requires a recent browser): > > <https://jsfiddle.net/juliopdiegidio/aqxfLvs2/embedded/result/> > > It is a symmetric setup, only the radial motion is considered. You must mean the tangential motion. > The first diagram and figures shows the point of view of a C that > stays in the mid-point between A and B. Quite. The whole scenario is observed in the rest frame of the inertial C, that is in an inertial frame. So the solution is trivial. > > The second diagram and figures shows the point of view of A. No it doesn't. Not even close! What you have missed is that A isn't inertial, he is accelerating. Observing an accelerating twin in an accelerated frame of reference is quite complicated. > The two points of view (of C and of A) turn out to be incompatible: > there is a rendez-vous between all three observes from the point of > view of C, but no such rendez-vous from the point of view of A (or, symmetrically, for B). >-- But I might also have made some basic mistake. Indeed you have. > Anybody here knows how to fix it ?? > > Julio > I will not do the complex job of calculating your scenario. I can however show you a simulation where the "twin paradox" is viewed from both twins' points of view. And yes, the views are compatible. https://paulba.no/twins.html -- Paul https://paulba.no/
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| From | Julio Di Egidio <julio@diegidio.name> |
|---|---|
| Date | 2016-04-03 15:13 -0700 |
| Message-ID | <055f2999-aa64-4a3b-b10a-2bddcab6bc92@googlegroups.com> |
| In reply to | #380414 |
On Tuesday, March 29, 2016 at 1:54:40 PM UTC+1, Paul B. Andersen wrote:
> On 29.03.2016 13:40, Julio Di Egidio wrote:
>
> > Here is another little "experiment" (requires a recent browser):
> >
> > <https://jsfiddle.net/juliopdiegidio/aqxfLvs2/embedded/result/>
> >
> > It is a symmetric setup, only the radial motion is considered.
>
> You must mean the tangential motion.
Yes, of course.
> > The second diagram and figures shows the point of view of A.
>
> No it doesn't. Not even close!
> What you have missed is that A isn't inertial, he is accelerating.
Nope, that is not the problem. This is simply collinear motion, just think of it as flat spacetime that is finitely periodic, i.e. when you exit from one edge, you re-enter from the opposite edge. And I'd insist that is fine and all inertial... My mistakes, if any, as I am still looking for them, are elsewhere.
Incidentally, that you cannot set up an interesting special relativity problem without accelerated frames is your mistake: indeed, the peculiar properties of SR are, and, for pedagogical reason, should be made already apparent with just inertial frames.
> I will not do the complex job of calculating your scenario.
Following are the logic and calculations that I am using for the above "experiment". Please verify or shut up:
=========================================================================
Set the speed of light c = 1.
Consider three particles, A, C, and B in collinear motion.
To each particle associate, with the same name, a frame of reference
where the particle is at rest at the origin.
Assume the three particles have synchronised their clocks in such a way
that, at time 0 for each particle, the three particles are at the same
point in space.
Now consider this specific case:
In frame C, where C is at rest at the origin, let A be moving towards
the left of C at some constant speed -v, and let B be moving towards
the right of C at the opposite constant speed v.
Question 1:
-----------
When in frame C the time is t_{C} = 1, what is the time t_{A} in
frame A?
Answer 1:
t_{A}(t_{C}) = alpha(v) * t_{C}
t_{A}(1) = alpha(v)
Question 2:
-----------
In frame A, what are the speeds of C and B?
Answer 2:
The speed of C relative to A is the opposite of the speed of A
relative to C:
v_C_{A} = -v_A_{C} = -(-v) = v
The speed of B relative to A is the speed of C relative to A composed
with the speed of B relative to C:
v_B_{A} = (v_C_{A} + v_B_{C}) / (1 + (v_C_{A} * v_B_{C})) =
= (v + v) / (1 + (v * v)) =
= 2 * v / (1 + v^2)
Question 3:
-----------
In frame C, what are the positions of A, C, and B at time t = 1?
Answer 3:
In frame C (omitting '_{C}'):
x_A(t) = x_A(0) + v_A * t = -v * t
x_C(t) = x_C(0) + v_C * t = 0
x_B(t) = x_B(0) + v_B * t = v * t
x_A(1) = -v
x_C(1) = 0
x_B(1) = v
Question 4:
-----------
In frame A, what are the positions of A, C and B at time t = alpha(v)
(i.e. when it is t = 1 in frame C)?
Answer 4:
In frame A (omitting '_{A}'):
x_A(t) = x_A(0) + v_A * t = 0
x_C(t) = x_C(0) + v_C * t = v * t
x_B(t) = x_B(0) + v_B * t = 2 * v * t / (1 + v^2)
x_A(alpha(v)) = 0
x_C(alpha(v)) = alpha(v) * v
x_B(alpha(v)) = alpha(v) * 2 * v / (1 + v^2)
=========================================================================
Correct so far?
Julio
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| From | moroney@world.std.spaamtrap.com (Michael Moroney) |
|---|---|
| Date | 2016-04-04 01:03 +0000 |
| Message-ID | <ndsehd$snp$2@pcls7.std.com> |
| In reply to | #380847 |
Julio Di Egidio <julio@diegidio.name> writes: >On Tuesday, March 29, 2016 at 1:54:40 PM UTC+1, Paul B. Andersen wrote: >> > The second diagram and figures shows the point of view of A. >> >> No it doesn't. Not even close! >> What you have missed is that A isn't inertial, he is accelerating. >Nope, that is not the problem. It most certainly is. The "stationary" observer is constantly accelerating as it is moving circularly. While position and velocity in SR are relative, acceleration is not. > This is simply collinear motion, It most certainly is not! The "stationary" observer is under constant acceleration by moving circularly.
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| From | David Waite <waitedavid1618@yahoo.com> |
|---|---|
| Date | 2016-04-03 20:57 -0700 |
| Message-ID | <1481cf6c-98ce-438e-9b10-ddd93c836ede@googlegroups.com> |
| In reply to | #380847 |
On Sunday, April 3, 2016 at 3:13:55 PM UTC-7, Julio Di Egidio wrote:
> On Tuesday, March 29, 2016 at 1:54:40 PM UTC+1, Paul B. Andersen wrote:
> > On 29.03.2016 13:40, Julio Di Egidio wrote:
> >
> > > Here is another little "experiment" (requires a recent browser):
> > >
> > > <https://jsfiddle.net/juliopdiegidio/aqxfLvs2/embedded/result/>
> > >
> > > It is a symmetric setup, only the radial motion is considered.
> >
> > You must mean the tangential motion.
>
> Yes, of course.
>
> > > The second diagram and figures shows the point of view of A.
> >
> > No it doesn't. Not even close!
> > What you have missed is that A isn't inertial, he is accelerating.
>
> Nope, that is not the problem. This is simply collinear motion, just think of it as flat spacetime that is finitely periodic, i.e. when you exit from one edge, you re-enter from the opposite edge. And I'd insist that is fine and all inertial... My mistakes, if any, as I am still looking for them, are elsewhere.
>
> Incidentally, that you cannot set up an interesting special relativity problem without accelerated frames is your mistake: indeed, the peculiar properties of SR are, and, for pedagogical reason, should be made already apparent with just inertial frames.
>
> > I will not do the complex job of calculating your scenario.
>
> Following are the logic and calculations that I am using for the above "experiment". Please verify or shut up:
>
> =========================================================================
>
> Set the speed of light c = 1.
>
> Consider three particles, A, C, and B in collinear motion.
>
> To each particle associate, with the same name, a frame of reference
> where the particle is at rest at the origin.
>
> Assume the three particles have synchronised their clocks in such a way
> that, at time 0 for each particle, the three particles are at the same
> point in space.
>
> Now consider this specific case:
>
> In frame C, where C is at rest at the origin, let A be moving towards
> the left of C at some constant speed -v, and let B be moving towards
> the right of C at the opposite constant speed v.
>
> Question 1:
> -----------
>
> When in frame C the time is t_{C} = 1, what is the time t_{A} in
> frame A?
>
> Answer 1:
>
> t_{A}(t_{C}) = alpha(v) * t_{C}
>
> t_{A}(1) = alpha(v)
>
> Question 2:
> -----------
>
> In frame A, what are the speeds of C and B?
>
> Answer 2:
>
> The speed of C relative to A is the opposite of the speed of A
> relative to C:
>
> v_C_{A} = -v_A_{C} = -(-v) = v
>
> The speed of B relative to A is the speed of C relative to A composed
> with the speed of B relative to C:
>
> v_B_{A} = (v_C_{A} + v_B_{C}) / (1 + (v_C_{A} * v_B_{C})) =
> = (v + v) / (1 + (v * v)) =
> = 2 * v / (1 + v^2)
>
> Question 3:
> -----------
>
> In frame C, what are the positions of A, C, and B at time t = 1?
>
> Answer 3:
>
> In frame C (omitting '_{C}'):
>
> x_A(t) = x_A(0) + v_A * t = -v * t
> x_C(t) = x_C(0) + v_C * t = 0
> x_B(t) = x_B(0) + v_B * t = v * t
>
> x_A(1) = -v
> x_C(1) = 0
> x_B(1) = v
>
> Question 4:
> -----------
>
> In frame A, what are the positions of A, C and B at time t = alpha(v)
> (i.e. when it is t = 1 in frame C)?
>
> Answer 4:
>
> In frame A (omitting '_{A}'):
>
> x_A(t) = x_A(0) + v_A * t = 0
> x_C(t) = x_C(0) + v_C * t = v * t
> x_B(t) = x_B(0) + v_B * t = 2 * v * t / (1 + v^2)
>
> x_A(alpha(v)) = 0
> x_C(alpha(v)) = alpha(v) * v
> x_B(alpha(v)) = alpha(v) * 2 * v / (1 + v^2)
>
> =========================================================================
>
> Correct so far?
>
> Julio
Dumb nut, I already completely invalidated you here,
I know this is over your crackpot head, but I’ll suppose I’ll spell it out anyway. Lets say Al is an unaccelerated observer in the flat spacetime of special relativity. Coordinates appropriate for an inertial frame observer express the spacetime line element as
ds²=dct²-dx²-dy²-dz²
Al’s inertial frame coordinates will be t,x,y,z and that will be the expression for the line element according to his coordinates.
Now consider another observer Bert, who may even be accelerated.
Local to him using locally rectilinear coordinates he will find that the line element reduces to
ds²→ dct’²-dx’²-dy’²-dz’²
He doesn’t move with respect to himself so using this as a description of his own path through spacetime the spatial differentials vanish and the path describes his wristwatch time, proper time.
ds²→ dcτ²
Since ds² is an invariant we see that the path of Bert as described by Al’s inertial frame coordinates is
dct²-dx²-dy²-dz²= dcτ²
Algebra
dct²{1-[(dx/dct)²+(dy/dct)²+(dz/dct)²]}= dcτ²
dt²{1-[(dx/dct)²+(dy/dct)²+(dz/dct)²]}= dτ²
dt²[1-(v²/c²)]= dτ²
dt²= dτ²/[1-(v²/c²)]
dt= dτ/sqrt[1-(v²/c²)]
So this time dilation equation is how the inertial frame observer Al, observes the other observer Bert to be time dilated, no matter whether Bert is accelerated or not.
Now Julio is making the crackpot assertion that even when accelerated Bert would get the result finding Al time dilated. Relativity says he doesn’t and relativity yields no contradiction, so claiming otherwise is a strawman fallacy. To find how things go according to the accelerated spacetime standards of Bert, you must first assign Bert accelerated coordinates appropriate to his accelerated state. So lets say Bert accelerates along the x-axis. We’ll assign him accelerated coordinate of t’ and x’ according to
ct=∫γdct'+γβx'
x=γx'+∫γβdct'
where γ and β are to be expressed as functions of t’. Now he accelerates with arbitrary time dependence but two things can clearly be seen before we even go any further. First of all, these reduce to Lorentz transformation only when he is not accelerating, and only then will he observe Al to be time dilated, however, second of all, since he is always at x’=0, he has to agree with Al that in any round trip when the watches are compared side be side that it will be his own watch that will have recorded less time in accordance with special relativistic time dilation of
Δct=∫γdct'
as that comes directly from the first transformation equation for the coordinates assigned him.
Now under this transformation the line element expressed in accelerated Bert’s coordinates is
ds²=(1+αx'/c²)²dct'²-dx'²-dy²-dz²
α=γ²(dβ/dt')c
From a general relativity perspective, what Bert describes as going on is he describes inertial Al’s motion as a geodesic in what he perceives as a gravitational acceleration field which yields a gravitational red or blue shift or gravitational time dilation effect on clocks far along the x’ direction corresponding to the (1+αx'/c²)² element of the metric. According to the accelerated spacetime standards appropriate for Bert, he finds that when Al is far away, and he accelerates toward Al, Al’s watch runs fast corresponding to a blue shift. In the end both observers agree that after a round trip, Burt aged less and by how much.
Now Julio just introduces another accelerated frame. It doesn’t matter. You still have to treat that frame as accelerated, the same way we had to treat Bert’s frame as accelerated and assigned him appropriate accelerated coordinates. The kind of acceleration that Bert experiences is absolute. It is not relative and is not treated as relative in relativity. Saying otherwise is a strawman fallacy. You can tell who accelerates and in the end all observers can agree on that. If the spacetime you observe is
ds²=dct²-dx²-dy²-dz²
then you’re not accelerating.
If the spacetime you observe is something like
ds²=(1+αx'/c²)²dct'²-dx'²-dy²-dz²
then you absolutely ARE accelerating.
The spacetime length of the acceleration 4-vector on you in that case is nonzero, and that is an invariant,
so why are you spamming this thread?
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| From | Koobee Wublee <koobee.wublee@gmail.com> |
|---|---|
| Date | 2016-04-03 22:14 -0700 |
| Message-ID | <4e652c90-16d6-4b52-95a9-0d542d4693b0@googlegroups.com> |
| In reply to | #380857 |
On Sunday, April 3, 2016 at 8:57:36 PM UTC-7, David Waite wrote: > Lets say Al is an unaccelerated observer... > ds²→ dct’²-dx’²-dy’²-dz’² This equation above does not give a fvck if an observer is accelerated or not. <shrug> > Since ds² is an invariant... ds² is invariant because it is (c² dτ²) where dτ is the local rate of time flow that is independent of whoever is observing whoever is in this local time. <shrug> > dt= dτ/sqrt[1-(v²/c²)] . . . (1) Yes, the equation of Minkowski spacetime directly arrive at the equation above. <shrug> > So this time dilation equation is... <nonsense snipped> dτ is the local time at wherever it is observed by a single or multiple observers. Of course, this local time is invariant because no observer can affect whatever the tick rate of time flow at this very remote location. <shrug> > ct=∫γdct'+γβx' ... <more nonsense> To find out the time elapse, all you have to do is to integrate equation 1 which is: ** ∫(dτ) = ∫(dt sqrt(1 - v²/c²)) Remember ** dτ = invariant local rate of time flow at the observed ** dt = observer’s rate of time flow ** v = speed of the observed relative to the observer dt Placing the boundary conditions to the equation above, time is reset when the twins were at rest with each other. The complete equation is: ** ∫[0, Tτ](dτ) = ∫[0, Tt](dt sqrt(1 - v²/c²)) Where ** Tτ = calendar time of the observed ** Tt = calendar time of the observer Now, say the twins are A and B, the equation above would spawn two separate, independent equations as described below. <shrug> ** ∫[0, TA](dτ) = ∫[0, TB](dt sqrt(1 - v²/c²)) . . . (2) And ** ∫[0, TB](dτ) = ∫[0, TA](dt sqrt(1 - v²/c²)) . . . (3) Where ** Equation (2) describes B observing A, and (3) describes A observing B. There are no known values for TA and TB that are arithmeticly consistent unless (TA = TB = 0) or [(v = 0) and (TA = TB)]. Thus, the Twin paradox represents a fatal contradiction to SR. How can idiots argue against this simple mathematics? <shrug>
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| From | David Waite <waitedavid1618@yahoo.com> |
|---|---|
| Date | 2016-04-04 10:53 -0700 |
| Message-ID | <672024d0-6817-4524-a530-17817198f524@googlegroups.com> |
| In reply to | #380859 |
On Sunday, April 3, 2016 at 10:14:47 PM UTC-7, Koobee Wublee wrote: > On Sunday, April 3, 2016 at 8:57:36 PM UTC-7, David Waite wrote: > > > Lets say Al is an unaccelerated observer... > > ds²→ dct’²-dx’²-dy’²-dz’² > > This equation above does not give a fvck if an observer is accelerated or not. <shrug> > > > Since ds² is an invariant... > > ds² is invariant because it is (c² dτ²) where dτ is the local rate of time flow that is independent of whoever is observing whoever is in this local time. <shrug> > > > dt= dτ/sqrt[1-(v²/c²)] . . . (1) > > Yes, the equation of Minkowski spacetime directly arrive at the equation above. <shrug> > > > So this time dilation equation is... > > <nonsense snipped> dτ is the local time at wherever it is observed by a single or multiple observers. Of course, this local time is invariant because no observer can affect whatever the tick rate of time flow at this very remote location. <shrug> > > > ct=∫γdct'+γβx' ... > > <more nonsense> > > To find out the time elapse, all you have to do is to integrate equation 1 which is: > > ** ∫(dτ) = ∫(dt sqrt(1 - v²/c²)) > > Remember > > ** dτ = invariant local rate of time flow at the observed > ** dt = observer’s rate of time flow > ** v = speed of the observed relative to the observer dt > > Placing the boundary conditions to the equation above, time is reset when the twins were at rest with each other. The complete equation is: > > ** ∫[0, Tτ](dτ) = ∫[0, Tt](dt sqrt(1 - v²/c²)) > > Where > > ** Tτ = calendar time of the observed > ** Tt = calendar time of the observer > > Now, say the twins are A and B, the equation above would spawn two separate, independent equations as described below. <shrug> > > ** ∫[0, TA](dτ) = ∫[0, TB](dt sqrt(1 - v²/c²)) . . . (2) > > And > > ** ∫[0, TB](dτ) = ∫[0, TA](dt sqrt(1 - v²/c²)) . . . (3) > > Where > > ** Equation (2) describes B observing A, and (3) describes A observing B. > > There are no known values for TA and TB that are arithmeticly consistent unless (TA = TB = 0) or [(v = 0) and (TA = TB)]. Thus, the Twin paradox represents a fatal contradiction to SR. How can idiots argue against this simple mathematics? <shrug> Dumb nut, I already debunked you as a crank, https://groups.google.com/forum/#!topic/sci.physics.relativity/KbPmC4vUDXI Since I've already shown that you don't actually understand the subject, you have nothing to contribute.
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| From | David Waite <waitedavid1618@yahoo.com> |
|---|---|
| Date | 2016-04-04 11:24 -0700 |
| Message-ID | <3a99c24d-31b5-4ba0-86ad-0303541d0d32@googlegroups.com> |
| In reply to | #380859 |
On Sunday, April 3, 2016 at 10:14:47 PM UTC-7, Koobee Wublee wrote: > On Sunday, April 3, 2016 at 8:57:36 PM UTC-7, David Waite wrote: > > We’ll assign him accelerated coordinate of t’ and x’ according to > > ct=∫γdct'+γβx' > > x=γx'+∫γβdct' > > where γ and β are to be expressed as functions of t’. > > <more nonsense> Since Koobee just proved coordinate transformations are over his head, and I already knew it would be over the heads of such rather unintelligent cranks, I didn't expect it to sink in. Explaining relativity to Herbert Dingle worshipers, i.e. Dingleberrys, is like trying to explain Newtonian mechanics to my cat. However there are noncranks that may come across the thread with an interest in learning so I'll give an example how to apply this for their sake. Lets say the accelerated primed frame observer has a velocity as a function of his time of v=ctanh(αt'/c) where α is a constant Then you get β=tanh(αt'/c) and with some simple math you will get γ=cosh(αt'/c) and γβ=sinh(αt'/c) So we then assign the accelerated primed frame observer coordinates related to the unprimed coordinates by ct=∫γdct'+γβx' x=γx'+∫γβdct' and inserting from above you get ct=c∫cosh(αt'/c)dt'+x'sinh(αt'/c) x=x'cosh(αt'/c)+c∫sinh(αt'/c)dt' Doing the anti-derivatives you get ct=(c²/α)sinh(αt'/c)+x'sinh(αt'/c)+k1 x=x'cosh(αt'/c)+(c²/α)cosh(αt'/c)+k2 We'll choose constants so that origins coincide when their times are zero ct=[(c²/α)+x']sinh(αt'/c) x=[(c²/α)+x']cosh(αt'/c)-(c²/α) Though the anti-derivative form of the transformation for arbitrarily primed time dependent acceleration ct=∫γdct'+γβx' x=γx'+∫γβdct' are more general, the constant proper acceleration case yielding what we just found ct=[(c²/α)+x']sinh(αt'/c) x=[(c²/α)+x']cosh(αt'/c)-(c²/α) gives Rindler coordinates for the primed frame. Look up Rindler coordinates.
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| From | Koobee Wublee <koobee.wublee@gmail.com> |
|---|---|
| Date | 2016-04-04 23:44 -0700 |
| Message-ID | <78c63436-b791-4479-8730-faddc5632eac@googlegroups.com> |
| In reply to | #380902 |
On Monday, April 4, 2016 at 11:24:35 AM UTC-7, David Waite wrote: > On Sunday, April 3, 2016 at 10:14:47 PM UTC-7, Koobee Wublee wrote: > > On Sunday, April 3, 2016 at 8:57:36 PM UTC-7, David Waite wrote: > > > We’ll assign him accelerated coordinate of t’ and x’ according to > > > ct=∫γdct'+γβx' > > > x=γx'+∫γβdct' > > > where γ and β are to be expressed as functions of t’. > > > > <more nonsense> > > Since Koobee just proved coordinate transformations are over his head, and I already knew it would be over the heads of such rather unintelligent cranks, I didn't expect it to sink in. Explaining relativity to Herbert Dingle worshipers, i.e. Dingleberrys, is like trying to explain Newtonian mechanics to my cat. However there are noncranks that may come across the thread with an interest in learning so I'll give an example how to apply this for their sake. Lets say the accelerated primed frame observer has a velocity as a function of his time of > v=ctanh(αt'/c) > where α is a constant > Then you get > β=tanh(αt'/c) > and with some simple math you will get > γ=cosh(αt'/c) > and > γβ=sinh(αt'/c) > So we then assign the accelerated primed frame observer coordinates related to the unprimed coordinates by > ct=∫γdct'+γβx' > x=γx'+∫γβdct' > and inserting from above you get > ct=c∫cosh(αt'/c)dt'+x'sinh(αt'/c) > x=x'cosh(αt'/c)+c∫sinh(αt'/c)dt' > Doing the anti-derivatives you get > ct=(c²/α)sinh(αt'/c)+x'sinh(αt'/c)+k1 > x=x'cosh(αt'/c)+(c²/α)cosh(αt'/c)+k2 > We'll choose constants so that origins coincide when their times are zero > ct=[(c²/α)+x']sinh(αt'/c) > x=[(c²/α)+x']cosh(αt'/c)-(c²/α) > Though the anti-derivative form of the transformation for arbitrarily primed time dependent acceleration > ct=∫γdct'+γβx' > x=γx'+∫γβdct' > are more general, the constant proper acceleration case yielding what we just found > ct=[(c²/α)+x']sinh(αt'/c) > x=[(c²/α)+x']cosh(αt'/c)-(c²/α) > gives Rindler coordinates for the primed frame. > Look up Rindler coordinates. Idiot²!!!
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| From | David Waite <waitedavid1618@yahoo.com> |
|---|---|
| Date | 2016-04-05 02:05 -0700 |
| Message-ID | <7968f678-4043-43bb-8401-1adeadf40284@googlegroups.com> |
| In reply to | #380928 |
On Monday, April 4, 2016 at 11:44:18 PM UTC-7, Koobee Wublee wrote:
> On Monday, April 4, 2016 at 11:24:35 AM UTC-7, David Waite wrote:
> > On Sunday, April 3, 2016 at 10:14:47 PM UTC-7, Koobee Wublee wrote:
> > > On Sunday, April 3, 2016 at 8:57:36 PM UTC-7, David Waite wrote:
> > > > We’ll assign him accelerated coordinate of t’ and x’ according to
> > > > ct=∫γdct'+γβx'
> > > > x=γx'+∫γβdct'
> > > > where γ and β are to be expressed as functions of t’.
> > >
> > > <more nonsense>
> >
> > Since Koobee just proved coordinate transformations are over his head, and I already knew it would be over the heads of such rather unintelligent cranks, I didn't expect it to sink in. Explaining relativity to Herbert Dingle worshipers, i.e. Dingleberrys, is like trying to explain Newtonian mechanics to my cat. However there are noncranks that may come across the thread with an interest in learning so I'll give an example how to apply this for their sake. Lets say the accelerated primed frame observer has a velocity as a function of his time of
> > v=ctanh(αt'/c)
> > where α is a constant
> > Then you get
> > β=tanh(αt'/c)
> > and with some simple math you will get
> > γ=cosh(αt'/c)
> > and
> > γβ=sinh(αt'/c)
> > So we then assign the accelerated primed frame observer coordinates related to the unprimed coordinates by
> > ct=∫γdct'+γβx'
> > x=γx'+∫γβdct'
> > and inserting from above you get
> > ct=c∫cosh(αt'/c)dt'+x'sinh(αt'/c)
> > x=x'cosh(αt'/c)+c∫sinh(αt'/c)dt'
> > Doing the anti-derivatives you get
> > ct=(c²/α)sinh(αt'/c)+x'sinh(αt'/c)+k1
> > x=x'cosh(αt'/c)+(c²/α)cosh(αt'/c)+k2
> > We'll choose constants so that origins coincide when their times are zero
> > ct=[(c²/α)+x']sinh(αt'/c)
> > x=[(c²/α)+x']cosh(αt'/c)-(c²/α)
> > Though the anti-derivative form of the transformation for arbitrarily primed time dependent acceleration
> > ct=∫γdct'+γβx'
> > x=γx'+∫γβdct'
> > are more general, the constant proper acceleration case yielding what we just found
> > ct=[(c²/α)+x']sinh(αt'/c)
> > x=[(c²/α)+x']cosh(αt'/c)-(c²/α)
> > gives Rindler coordinates for the primed frame.
> > Look up Rindler coordinates.
>
> Idiot²!!!
So,
Though the anti-derivative form of the transformation for arbitrarily primed time dependent acceleration
ct=∫γdct'+γβx'
x=γx'+∫γβdct'
are more general, the constant proper acceleration case yielding what we just found
ct=[(c²/α)+x']sinh(αt'/c)
x=[(c²/α)+x']cosh(αt'/c)-(c²/α)
gives Rindler coordinates for the primed frame.
Look up Rindler coordinates,
was over your head. Big surprise. As I said, I already knew you can't understand such simple things as such coordinate transformations. Anyway, moving on for the noncranks who can learn it, lets look at the expression for the line element and do the transformation
Differentiating the transformation you get
dct=γdct'+d(γβx')
dx=d(γx')+γβdct'
dct=γdct'+γβdx'+γx'(dβ/dct')dct'+βx'(dγ/dct')dct'
dx=γβdct'+γdx'+x'd(γ/dct')dct'
dct=γdct'+γβdx'+γx'(dβ/dct')dct'+γ³β²x'(dβ/dct')dct'
dx=γβdct'+γdx'+γ³βx'(dβ/dct')dct'
dct=γdct'+γx'(dβ/dct')dct'+γ³β²x'(dβ/dct')dct'+γβdx'
dx=γβdct'+γ³βx'(dβ/dct')dct'+γdx'
dct= γ[1+x'(dβ/dct')+γ²β²x'(dβ/dct')]dct'+γβdx'
dx= γβ [1+γ²x'(dβ/dct')]dct'+γdx'
dct= γ[1+ γ²(1- β²)x'(dβ/dct')+γ²β²x'(dβ/dct')]dct'+γβdx'
dx= γβ [1+γ²x'(dβ/dct')]dct'+γdx'
dct= γ[1+ γ²x'(dβ/dct')]dct'+γβdx'
dx= γβ [1+γ²x'(dβ/dct')]dct'+γdx'
Define α
α=cγ²dβ/dt'
dct= γ[1+(αx'/c²)]dct'+γβdx'
dx= γβ [1+(αx'/c²)]dct'+γdx'
square them
dct²= γ²[1+(αx'/c²)]²dct'²+2γ²β[1+(αx'/c²)]dct'dx’'+γ²β²dx'²
dx= γ²β²[1+(αx'/c²)]²dct'²+2γ²β[1+(αx'/c²)]dct'dx’'+γ²dx'²
Insert into
ds²=dct²-dx²-dy²-dz²
you get
ds²= γ²[1+(αx'/c²)]²dct'²+2γ²β[1+(αx'/c²)]dct'dx’'+γ²β²dx'²-{γ²β²[1+(αx'/c²)]²dct'²+2γ²β[1+(αx'/c²)]dct'dx’'+γ²dx'²}-dy²-dz²
ds²= γ²[1+ (αx'/c²)]²dct'²+2γ²β[1+(αx'/c²)]dct'dx’'+γ²β²dx'²-γ²β²[1+(αx'/c²)]²dct'²-2γ²β[1+(αx'/c²)]dct'dx’'-γ²dx'²-dy²-dz²
ds²= γ²[1+(αx'/c²)]²dct'²+γ²β²dx'²-γ²β²[1+(αx'/c²)]²dct'²-γ²dx'²-dy²-dz²
ds²= (γ²-γ²β²)[1+(αx'/c²)]²dct'²-(γ²-γ²β²)dx'²-dy²-dz²
ds²= γ²(1-β²)[1+(αx'/c²)]²dct'²- γ²(1-β²)dx'²-dy²-dz²
ds²= [1+(αx'/c²)]²dct'²-dx'²-dy²-dz²
So that is how the spacetime is expressed according to the accelerated observer
ds²= [1+(αx'/c²)]²dct'²-dx'²-dy²-dz²
Now Koobee, did I include enough of the simple math steps along the way or was it still over your head? Oh wait, I already know the answer to that, simple math is over your head.
Anyway generalizing a little further still
ds²= [1+(α₁x'/c²)+(α₂y'/c²)+(α₃z'/c²)]²dct'²-dx'²-dy'²-dz'²
Where α₁, α₂, α₃ , are arbitrary functions of time t' corresponds to a zero Reimann tensor and as such is an appropriate coordinate expression of spacetime according to an arbitrarily accelerated observer in arbitrary directions with arbitrary time dependence in otherwise flat spacetime.
And the point as I was saying is that the accelerated observer agrees with the unaccelerated observer that it is his own watch that records less time in accord with special relativity and there is no such paradox in the "twin paradox" when the problem is worked correctly like this. The accelerated observer using these coordinates merely interprets the result as a gravitational doppler shift on remote clocks that occurs when he accelerates.
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| From | mlwozniak@wp.pl |
|---|---|
| Date | 2016-04-05 02:51 -0700 |
| Message-ID | <5c333c9d-8b1a-4172-90f4-96b67487912d@googlegroups.com> |
| In reply to | #380940 |
W dniu wtorek, 5 kwietnia 2016 11:06:01 UTC+2 użytkownik David Waite napisał: > And the point as I was saying is that the accelerated observer agrees with the unaccelerated observer that it is his own watch that records less time in accord with special relativity And observer walking a street agrees that trees, building and lanterns are running around him. So said The Shit!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!! Must be the truth!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
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| From | David Waite <waitedavid1618@yahoo.com> |
|---|---|
| Date | 2016-04-05 02:58 -0700 |
| Message-ID | <b33826d2-25d5-4672-9640-11a90539c242@googlegroups.com> |
| In reply to | #380941 |
On Tuesday, April 5, 2016 at 2:51:10 AM UTC-7, mlwo...@wp.pl wrote: > W dniu wtorek, 5 kwietnia 2016 11:06:01 UTC+2 użytkownik David Waite napisał: > > > And the point as I was saying is that the accelerated observer agrees with the unaccelerated observer that it is his own watch that records less time in accord with special relativity > > And observer walking a street agrees that trees, building > and lanterns are running around him. > So said The Shit!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!! > Must be the truth!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!! Yeah yeah we already know that your an Aristotelian thinker, a drooling moron that doesn't get Galilean relativity. Oh wait, you're such a moron that you didn't even know that your argument was against Galilean relativity. Oh wait, you're such a drooling moron that you don't even know what Galilean relativity means and were just speculating that you were addressing special relativity.
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| From | mlwozniak@wp.pl |
|---|---|
| Date | 2016-04-05 04:08 -0700 |
| Message-ID | <4ee21039-0fa5-422b-ace1-4036fcd0d049@googlegroups.com> |
| In reply to | #380942 |
W dniu wtorek, 5 kwietnia 2016 11:58:36 UTC+2 użytkownik David Waite napisał: > On Tuesday, April 5, 2016 at 2:51:10 AM UTC-7, mlwo...@wp.pl wrote: > > W dniu wtorek, 5 kwietnia 2016 11:06:01 UTC+2 użytkownik David Waite napisał: > > > > > And the point as I was saying is that the accelerated observer agrees with the unaccelerated observer that it is his own watch that records less time in accord with special relativity > > > > And observer walking a street agrees that trees, building > > and lanterns are running around him. > > So said The Shit!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!! > > Must be the truth!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!! > > Yeah yeah we already know that your an Aristotelian thinker, a drooling moron that doesn't get Galilean relativity. Oh wait, you're such a moron that you didn't even know that your argument was against Galilean relativity. Oh wait, you're such a drooling moron that you don't even know what Galilean relativity means and were just speculating that you were addressing special relativity. Yeah yeah we already know you're an relativistic thinker, a drooling moron completely disconnected from reality, with one and only argument "a GURU said!". Galileo didn't really believe his concept himself, you know? he also said: "and yet it's moving". The common sense was strong in this one.
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| From | David Waite <waitedavid1618@yahoo.com> |
|---|---|
| Date | 2016-04-05 04:25 -0700 |
| Message-ID | <164a5f2c-6576-4bbb-89b1-0aea52473ddc@googlegroups.com> |
| In reply to | #380943 |
On Tuesday, April 5, 2016 at 4:08:23 AM UTC-7, mlwo...@wp.pl wrote: > Galileo didn't really believe his concept himself, you know? As I said, you're such a moron that you're an Aristotelean thinker arguing not against Einstein actually, but against Galilean relativity. Got it.
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| From | mlwozniak@wp.pl |
|---|---|
| Date | 2016-04-05 04:41 -0700 |
| Message-ID | <25262e6c-8e11-4d6a-8a1b-d9a29e175db2@googlegroups.com> |
| In reply to | #380946 |
W dniu wtorek, 5 kwietnia 2016 13:25:16 UTC+2 użytkownik David Waite napisał: > On Tuesday, April 5, 2016 at 4:08:23 AM UTC-7, mlwo...@wp.pl wrote: > > Galileo didn't really believe his concept himself, you know? > > As I said, you're such a moron that you're an Aristotelean thinker arguing not against Einstein actually, but against Galilean relativity. Got it. Oh, really? Have you got it? after only about 20 or 30 direct claims? Great job, poor idiot. As I said, deep inside your guru was Aristotelean too. Samely, as Copernicus. "And yet it moves...", you know.
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| From | David Waite <waitedavid1618@yahoo.com> |
|---|---|
| Date | 2016-04-05 05:14 -0700 |
| Message-ID | <d7c8bb4e-55e5-4685-97d8-ec4a86556151@googlegroups.com> |
| In reply to | #380947 |
On Tuesday, April 5, 2016 at 4:41:33 AM UTC-7, mlwo...@wp.pl wrote: > W dniu wtorek, 5 kwietnia 2016 13:25:16 UTC+2 użytkownik David Waite napisał: > > On Tuesday, April 5, 2016 at 4:08:23 AM UTC-7, mlwo...@wp.pl wrote: > > > Galileo didn't really believe his concept himself, you know? > > > > As I said, you're such a moron that you're an Aristotelean thinker arguing not against Einstein actually, but against Galilean relativity. Got it. > > Oh, really? Have you got it? after only about 20 or > 30 direct claims? Great job, poor idiot. > As I said, deep inside your guru was Aristotelean too. > Samely, as Copernicus. "And yet it moves...", you know. So the ape is now pounding his fists on the keyboard and hoping something meaningful comes out of it. Get an education ape-man.
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| From | mlwozniak@wp.pl |
|---|---|
| Date | 2016-04-05 05:46 -0700 |
| Message-ID | <721f9e76-9660-4acb-b1c3-5828f00ec651@googlegroups.com> |
| In reply to | #380951 |
W dniu wtorek, 5 kwietnia 2016 14:15:00 UTC+2 użytkownik David Waite napisał: > On Tuesday, April 5, 2016 at 4:41:33 AM UTC-7, mlwo...@wp.pl wrote: > > W dniu wtorek, 5 kwietnia 2016 13:25:16 UTC+2 użytkownik David Waite napisał: > > > On Tuesday, April 5, 2016 at 4:08:23 AM UTC-7, mlwo...@wp.pl wrote: > > > > Galileo didn't really believe his concept himself, you know? > > > > > > As I said, you're such a moron that you're an Aristotelean thinker arguing not against Einstein actually, but against Galilean relativity. Got it. > > > > Oh, really? Have you got it? after only about 20 or > > 30 direct claims? Great job, poor idiot. > > As I said, deep inside your guru was Aristotelean too. > > Samely, as Copernicus. "And yet it moves...", you know. > > So the ape is now pounding his fists on the keyboard and hoping something meaningful comes out of it. Get an education ape-man. Rave, moron, and spit. What else can an ape do.
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| From | David Waite <waitedavid1618@yahoo.com> |
|---|---|
| Date | 2016-04-05 06:46 -0700 |
| Message-ID | <46625f2f-40d1-4b47-a5e9-c358cd07c927@googlegroups.com> |
| In reply to | #380953 |
On Tuesday, April 5, 2016 at 5:46:05 AM UTC-7, mlwo...@wp.pl wrote: > W dniu wtorek, 5 kwietnia 2016 14:15:00 UTC+2 użytkownik David Waite napisał: > > On Tuesday, April 5, 2016 at 4:41:33 AM UTC-7, mlwo...@wp.pl wrote: > > > W dniu wtorek, 5 kwietnia 2016 13:25:16 UTC+2 użytkownik David Waite napisał: > > > > On Tuesday, April 5, 2016 at 4:08:23 AM UTC-7, mlwo...@wp.pl wrote: > > > > > Galileo didn't really believe his concept himself, you know? > > > > > > > > As I said, you're such a moron that you're an Aristotelean thinker arguing not against Einstein actually, but against Galilean relativity. Got it. > > > > > > Oh, really? Have you got it? after only about 20 or > > > 30 direct claims? Great job, poor idiot. > > > As I said, deep inside your guru was Aristotelean too. > > > Samely, as Copernicus. "And yet it moves...", you know. > > > > So the ape is now pounding his fists on the keyboard and hoping something meaningful comes out of it. Get an education ape-man. > > Rave, moron, and spit. What else can an ape do. Yes I see that is all you do. So for sapiens, not you, lets recap. An unaccelerated observer of special relativity describes the spacetime by ds²=dct²-dx²-dy²-dz² From this we see that he observes that clocks in motion with respect to himself are time dilated according to dt= dτ/sqrt[1-(v²/c²)] whether or not that clock in motion with respect to himself is accelerated. However, an accelerated observer does not describe the spacetime as that, but as ds²=[1+(α₁x'/c²)+(α₂y'/c²)+(α₃z'/c²)]²dct'²-dx'²-dy'²-dz'² where α₁, α₂, α₃ , are arbitrary functions of time t'. So for example putting his time dependent acceleration along the x direction we may assign him coordinates related to the unaccelerated coordinates by ct=∫γdct'+γβx' x=γx'+∫γβdct' which transforms the expression ds²=dct²-dx²-dy²-dz² into ds²=[1+(αx'/c²)]²dct'²-dx'²-dy²-dz² The transformation is only Lorentz transformation when he is not accelerating. And for a round trip he has to agree with the inertial frame observer that it is his clock that recorded less time in accord with special relativity because for a round trip Δct=∫γdct' is directly yield by the transformation above. Thus there can't be an actual paradox in the "twin paradox". And we note that for constant proper acceleration the transformation became ct=[(c²/α)+x']sinh(αt'/c) x=[(c²/α)+x']cosh(αt'/c)-(c²/α) yielding Rindler coordinates as the primed accelerated observer coordinates. Just look up Rindler coordinates. And for the anti-relativity cranks, yes I know this is over your head and as an ape you will as you said, rave and spit as that's all a moronic ape like yourself can do.
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| From | Maciej Woźniak <mlwozniak@wp.pl> |
|---|---|
| Date | 2016-04-05 20:54 +0200 |
| Message-ID | <ne11mi$vih$1@node1.news.atman.pl> |
| In reply to | #380958 |
Użytkownik "David Waite" napisał w wiadomości grup dyskusyjnych:46625f2f-40d1-4b47-a5e9-c358cd07c927@googlegroups.com... On Tuesday, April 5, 2016 at 5:46:05 AM UTC-7, mlwo...@wp.pl wrote: > W dniu wtorek, 5 kwietnia 2016 14:15:00 UTC+2 użytkownik David Waite > napisał: > > On Tuesday, April 5, 2016 at 4:41:33 AM UTC-7, mlwo...@wp.pl wrote: > > > W dniu wtorek, 5 kwietnia 2016 13:25:16 UTC+2 użytkownik David Waite > > > napisał: > > > > On Tuesday, April 5, 2016 at 4:08:23 AM UTC-7, mlwo...@wp.pl wrote: > > > > > Galileo didn't really believe his concept himself, you know? > > > > > > > > As I said, you're such a moron that you're an Aristotelean thinker > > > > arguing not against Einstein actually, but against Galilean > > > > relativity. Got it. > > > > > > Oh, really? Have you got it? after only about 20 or > > > 30 direct claims? Great job, poor idiot. > > > As I said, deep inside your guru was Aristotelean too. > > > Samely, as Copernicus. "And yet it moves...", you know. > > > > So the ape is now pounding his fists on the keyboard and hoping > > something meaningful comes out of it. Get an education ape-man. > > Rave, moron, and spit. What else can an ape do. |Yes I see that is all you do. So for sapiens, not you, lets recap. An unaccelerated observer of special relativity describes the spacetime by |ds²=dct²-dx²-dy²-dz² |From this we see that he observes that clocks in motion with respect to himself are time dilated according to And observer walking a street obsrves that trees, building and lanterns are running around him. So said The Shit!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!! Must be the truth!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
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| From | David Waite <waitedavid1618@yahoo.com> |
|---|---|
| Date | 2016-04-05 12:45 -0700 |
| Message-ID | <8398a884-caca-4bde-9337-ea08bccff803@googlegroups.com> |
| In reply to | #380993 |
On Tuesday, April 5, 2016 at 11:55:15 AM UTC-7, Maciej Woźniak wrote: > Użytkownik "David Waite" napisał w wiadomości grup > dyskusyjnych:46625f2f-40d1-4b47-a5e9-c358cd07c927@googlegroups.com... > > On Tuesday, April 5, 2016 at 5:46:05 AM UTC-7, mlwo...@wp.pl wrote: > > W dniu wtorek, 5 kwietnia 2016 14:15:00 UTC+2 użytkownik David Waite > > napisał: > > > On Tuesday, April 5, 2016 at 4:41:33 AM UTC-7, mlwo...@wp.pl wrote: > > > > W dniu wtorek, 5 kwietnia 2016 13:25:16 UTC+2 użytkownik David Waite > > > > napisał: > > > > > On Tuesday, April 5, 2016 at 4:08:23 AM UTC-7, mlwo...@wp.pl wrote: > > > > > > Galileo didn't really believe his concept himself, you know? > > > > > > > > > > As I said, you're such a moron that you're an Aristotelean thinker > > > > > arguing not against Einstein actually, but against Galilean > > > > > relativity. Got it. > > > > > > > > Oh, really? Have you got it? after only about 20 or > > > > 30 direct claims? Great job, poor idiot. > > > > As I said, deep inside your guru was Aristotelean too. > > > > Samely, as Copernicus. "And yet it moves...", you know. > > > > > > So the ape is now pounding his fists on the keyboard and hoping > > > something meaningful comes out of it. Get an education ape-man. > > > > Rave, moron, and spit. What else can an ape do. > > |Yes I see that is all you do. So for sapiens, not you, lets recap. An > unaccelerated observer of special relativity describes the spacetime by > |ds²=dct²-dx²-dy²-dz² > |From this we see that he observes that clocks in motion with respect to > himself are time dilated according to > > And observer walking a street obsrves that trees, building > and lanterns are running around him. > So said The Shit!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!! > Must be the truth!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!! Yeah you already told us that you're such a complete moron that you're an Aristotelean thinker actually arguing against Galilean relativity, got it already. Back to the topic, an unaccelerated observer of special relativity describes the spacetime by ds²=dct²-dx²-dy²-dz² From this we see that he observes that clocks in motion with respect to himself are time dilated according to dt= dτ/sqrt[1-(v²/c²)] whether or not that clock in motion with respect to himself is accelerated. However, an accelerated observer does not describe the spacetime as that, but as ds²=[1+(α₁x'/c²)+(α₂y'/c²)+(α₃z'/c²)]²dct'²-dx'²-dy'²-dz'² where α₁, α₂, α₃ , are arbitrary functions of time t'. So for example putting his time dependent acceleration along the x direction we may assign him coordinates related to the unaccelerated coordinates by ct=∫γdct'+γβx' and x=γx'+∫γβdct' which transforms the expression ds²=dct²-dx²-dy²-dz² into ds²=[1+(αx'/c²)]²dct'²-dx'²-dy²-dz². The transformation is only Lorentz transformation when he is not accelerating. And for a round trip he has to agree with the inertial frame observer that it is his clock that recorded less time in accord with special relativity because for a round trip Δct=∫γdct' is directly yield by the transformation above. Thus there can't be an actual paradox in the "twin paradox". And we note that for constant proper acceleration the transformation became ct=[(c²/α)+x']sinh(αt'/c) and x=[(c²/α)+x']cosh(αt'/c)-(c²/α) yielding Rindler coordinates as the primed accelerated observer coordinates. Just look up Rindler coordinates. And for the anti-relativity cranks, yes I know this is over your head and as an ape you will as you said, rave and spit as that's all a moronic ape like yourself can do.
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